Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Explain solution RD Sharma class 12 chapter Straight Line in Space exercise 24.1 question 8 sub question (iv) maths

Explain solution RD Sharma class 12 chapter Straight Line in Space exercise 24.1 question 8 sub question (iv) maths

Answers (1)

Answer          :  4 \sqrt{2} \text { sq. units }

Hint                : To solve this equation we use area of parallelogram

Given              \hat{\imath}-3 \hat{\jmath}+\hat{k} \text { and } \hat{\imath}+\hat{\jmath}+\hat{k}

Solution          :\vec{a}=\hat{\imath}-3 \hat{\jmath}+\hat{k} \quad ; \quad \vec{b}=\hat{\imath}+\hat{\jmath}+\hat{k}

Area of parallelogram =|\vec{a} \times \vec{b}|

\begin{aligned} &\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 1 \\ 1 & 1 & 1 \end{array}\right| \\\\ &\vec{a} \times \vec{b}=\hat{i}[-3.1-1.1]-\hat{j}[1.1-1.1]+\hat{k}[1.1-1 \cdot(-3)] \end{aligned}

\begin{aligned} &=\hat{\mathrm{i}}[-3-1]-\hat{\mathrm{j}}[1-1]+\hat{\mathrm{k}}[1+3] \\\\ &\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=-4 \hat{\mathbf{i}}+4 \hat{\mathrm{k}} \end{aligned}

Area of parallelogram

\begin{aligned} &=\sqrt{-4^{2}+4^{2}} \\\\ &=4 \sqrt{2} \end{aligned}

Posted by

infoexpert26

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Date Sheet 2025 (Out) Live: Class 10, 12 board exam time table at cbse.gov.in; syllabus, sample papers
anam-khan

CBSE Class 12 board exam date sheet 2025 out; exams from February 15 to April 4
anam-khan

When will CBSE Board exam 2025 date sheet be released for Class 10, 12?

Latest Question